Hi again,
I assumed you didnt need instant response. If you need faster response then maybe current detection is the best bet.
To detect current we use Ohm's Law. We insert a resistor in series with the heater element and when the heater is 'on' we see a voltage drop across the resistor but when the heater is 'off' or blows open (as many element heaters do) that voltage goes to zero. That's the basic idea behind current sensing.
To sense the current you need to select the resistor value and the resistor power rating. the resistor value R comes from:
R=V/I
where
V is the required voltage drop in volts (to turn on the opto plus a little overhead),
I is the peak current of the heater element in amperes,
R is the required resistance in ohms.
The required safe power rating for the resistor would be:
Pr=2*I*I*R
where
I and R are defined above, and
Pr is the power rating in watts.
Lets look at an example. Say we have a heater that draws 2 amps and an opto coupler that has an internal LED that requires 1.7v to turn it on.
This means E=1.7 and I=2*1.4142=2.8 approximately. The required R would be:
R=E/I=1.7/2.8=0.6 ohms approximately.
The only catch is that we would like some overhead voltage so that we can also use a small value resistor in series with the opto to protect it against current surges in the line. Because of this we would like to increase the voltage by at least 0.5 volts. This will mean we'll need a little extra resistance which we'll call Ra. Now E=0.5 and I=2.8 so we have:
Ra=E/I=0.5/2.8=0.18 ohms approximately.
Now we add the two resistances to get R total:
RT=R+Ra=0.6+0.18=0.78 ohms total.
Thus the required resistor value should be around 0.8 ohms.
This causes heat because of the power, so we calculate a safe power for the resistor using the rms current of 2 amps (2.8 is the peak current):
Psafe=2*I*I*RT=2*2*2*0.8=6.4 watts,
so a 10 watt resistor would do nicely.
Next we need to calculate a small series resistance to protect the opto internal LED. To do this we use the voltage difference between the LED volts and the resistor voltage...
Vdiff=vRT-vLED
The peak voltage across RT will be RT*I=0.8*2.8=2.24 volts, so the difference will be:
Vdiff=vRT-vLED=2.24-1.70=0.54 volts.
Now the current into the opto will be roughly 5 to 10ma, so let's figure on 10ma for now. This makes the required resistor value:
Rs=Vdiff/iLED=0.54/0.010=54 ohms. We can use a 47 ohm resistor as that's a standard value.
Ok, so the entire circuit looks like a 0.8 ohm, 10 watt resistor in series with the heater element, with the opto input negative connected to one side of the resistor, and the other side of the resistor has the 47 ohm resistor connected to it and the other side of the 47 ohm resistor goes to the opto input positive connection.
The operation is such that for every half cycle, the opto turns on for a short period of time and then turns off. Thus you need to check the opto output for pulses and in the absence of pulses it will be assumed that the heater is off or disconnected, but possibly a brown out has occurred also (which means the heater is not working up to capacity).